HW06-solutions - wei(jw35975 – HW06 – kalahurka...

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Unformatted text preview: wei (jw35975) – HW06 – kalahurka – (55230) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Evaluate the integral I = integraldisplay 1 2 x- 10 x 2- x- 2 dx . 1. I = 6 ln2 correct 2. I =- 2 ln 3 3. I =- 6 ln 2 4. I =- 6 ln 3 5. I = 2 ln3 6. I = 2 ln2 Explanation: After factorization x 2- x- 2 = ( x + 1)( x- 2) . But then by partial fractions, 2 x- 10 x 2- x- 2 = 4 x + 1- 2 x- 2 . Now integraldisplay 1 4 x + 1 dx = bracketleftBig 4 ln | ( x + 1) | bracketrightBig 1 = 4 ln2 , while integraldisplay 1 2 x- 2 dx = bracketleftBig 2 ln | ( x- 2) | bracketrightBig 1 =- 2 ln2 . Consequently, I = 6 ln2 . 002 10.0 points Evaluate the definite integral I = integraldisplay 4 1 √ x x- 9 dx . 1. I = 3 ln 5 2- 2 2. I = 2- 3 ln 5 2 correct 3. I = 2- 3 ln5 4. I = 1- 3 ln5 5. I = 3 ln 5 2- 1 6. I = 3 ln5- 1 Explanation: Set u = √ x . Then u 2 = x, dx = 2 u du . In this case I = integraldisplay 2 1 u (2 u ) u 2- 9 du = integraldisplay 2 1 2 u 2 u 2- 9 du . The partial fraction decomposition is needed. First we divide: u 2 u 2- 9 = u 2- 9 + 9 u 2- 9 = 1 + 9 u 2- 9 = 1 + 9 ( u- 3)( u + 3) . But 1 ( u- 3)( u + 3) = 1 6 braceleftBig 1 u- 3- 1 u + 3 bracerightBig . Thus I = integraldisplay 2 1 braceleftBig 2 + 3 u- 3- 3 u + 3 bracerightBig du = bracketleftBig 2 u + 3 ln vextendsingle vextendsingle vextendsingle u- 3 u + 3 vextendsingle vextendsingle vextendsingle bracketrightBig 2 1 = 2 + 3 braceleftBig ln 1 5- ln 1 2 bracerightBig . wei (jw35975) – HW06 – kalahurka – (55230) 2 Consequently, I = 2- 3 ln 5 2 . 003 10.0 points Evaluate the integral I = integraldisplay 1 4 x 2 ( x + 1)( x 2 + 1) dx . 1. I = 2 parenleftBig π 2- ln 2 parenrightBig 2. I = ln 2 + π 2 3. I = 2 parenleftBig ln 2 + π 2 parenrightBig 4. I = 2 parenleftBig ln 8- π 2 parenrightBig 5. I = ln 8- π 2 correct 6. I = π 2- ln 2 Explanation: By partial fractions, 4 x 2 ( x + 1)( x 2 + 1) = A x + 1 + Bx + C x 2 + 1 . To determine A, B, and C multiply through by ( x + 1)( x 2 + 1): for then 4 x 2 = A ( x 2 + 1) + ( x + 1)( Bx + C ) = ( A + B ) x 2 + ( B + C ) x + ( A + C ) , which after comparing coefficients gives B =- C , A =- C , B = 2 . Thus I = 2 integraldisplay 1 parenleftBig 1 x + 1 + x- 1 x 2 + 1 parenrightBig dx = 2 parenleftBig integraldisplay 1 1 x + 1 dx + integraldisplay 1 x x 2 + 1 dx- integraldisplay 1 1 x 2 + 1 dx parenrightBig , and so I = 2 bracketleftBig ln( x + 1) + 1 2 ln( x 2 + 1)- tan- 1 x bracketrightBig 1 = bracketleftBig ln(( x + 1) 2 ( x 2 + 1))- 2 tan- 1 x bracketrightBig 1 . Consequently, I = ln 8- π 2 . 004 10.0 points Evaluate the definite integral I = integraldisplay 1 3 x 2- 5 x + 1 x 2- x- 2 dx ....
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This note was uploaded on 03/26/2012 for the course MATH 408 L taught by Professor Zheng during the Spring '10 term at University of Texas.

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HW06-solutions - wei(jw35975 – HW06 – kalahurka...

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